good afternoon. I have several doubts about how to change from one type of equation to another, and vice versa.
Suppose I have the following system generated in vector space:
$$\left \langle \begin{pmatrix} 5\\ 0\\ 2\\ 3\\ \end{pmatrix},\begin{pmatrix} 1\\ 3\\ 5\\ 1\\ \end{pmatrix} ,\begin{pmatrix} 1\\ 0\\ 0\\ 3\\ \end{pmatrix}, \begin{pmatrix} 0\\ 2\\ 0\\ 3\\ \end{pmatrix}\right \rangle \subset \mathbb{R}^4$$
- How can I obtain the parametric equations of the generator system?
- And how do I make these parametric equations implicit?
- Do they form a base? I think yes, because the rank of the matrix formed by the vectors of the generator system is maximum
Since you already proved that the four vectors form a basis of $\mathbb R^4$ by looking at the rank of the corresponding matrix, you can say that a system of parametric equations is just: $$\begin{cases} x_1 = a \\ x_2 = b \\ x_3 = c \\ x_4 = d \end{cases}$$ Indeed, the number of parameters is equal to the number of coordinates, because the subspace is equal to the whole space. A system of Cartesian equations would then be the empty system, because it needs to be satisfied by the coordinates of any vector.
Now, if you hadn't noticed this fact, you could have proceeded as follows.
Let $v = \begin{pmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{pmatrix}$ be a vector of the subspace. Then there exist $a, b, c, d \in \mathbb R$ such that $$v = a \begin{pmatrix} 5\\ 0\\ 2\\ 3\\ \end{pmatrix} + b \begin{pmatrix} 1\\ 3\\ 5\\ 1\\ \end{pmatrix} + c \begin{pmatrix} 1\\ 0\\ 0\\ 3\\ \end{pmatrix} + d\begin{pmatrix} 0\\ 2\\ 0\\ 3\\ \end{pmatrix}$$ i.e., such that $$\begin{pmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{pmatrix} = \begin{pmatrix} 5a+b+c\\ 3b+2d\\ 2a+5b\\ 3a+b+3c+3d\end{pmatrix}.$$ Therefore the parametric equations are: $$\begin{cases} x_1 = 5a+b+c \\ x_2 = 3b+2d \\ x_3 = 2a+5b \\ x_4 = 3a+b+3c+3d \end{cases}.$$
In order to obtain the Cartesian equations, you need to solve the linear system above for $a, b, c, d$. You can write it in matrix form as: $$\begin{pmatrix} 5 & 1 & 1 & 0 \\ 0 & 3 & 0 & 2 \\ 2 & 5 & 0 & 0 \\ 3 & 1 & 3 & 3 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}.$$ Then, by applying Gaussian elimination to the matrix $$\begin{pmatrix} 5 & 1 & 1 & 0 & x_1 \\ 0 & 3 & 0 & 2 & x_2 \\ 2 & 5 & 0 & 0 & x_3 \\ 3 & 1 & 3 & 3 & x_4 \end{pmatrix}$$ you get $$\begin{pmatrix} 1 & 0 & 0 & 0 & \frac 1 {94} (30 x_1 + 15 x_2 - 13 x_3 - 10 x_4) \\ 0 & 1 & 0 & 0 & \frac 1 {47} (-6 x_1 - 3 x_2 + 12 x_3 + 2 x_4) \\ 0 & 0 & 1 & 0 & \frac 1 {94} (-44 x_1 - 69 x_2 + 41 x_3 + 46 x_4) \\ 0 & 0 & 0 & 1 & \frac 1 {47} (9 x_1 + 28 x_2 - 18 x_3 - 3 x_4) \end{pmatrix}$$ hence there are no Cartesian equations, because the row echelon form already has leading $1$'s in all rows.
For the sake of giving you an example, if you removed the fourth vector the matrix would become: $$\begin{pmatrix} 5 & 1 & 1 & x_1 \\ 0 & 3 & 0 & x_2 \\ 2 & 5 & 0 & x_3 \\ 3 & 1 & 3 & x_4 \end{pmatrix}$$ which has row echelon form: $$\begin{pmatrix} 1 & 0 & 0 & \frac 1 6 (3 x_3 - 5 x_2) \\ 0 & 1 & 0 & \frac 1 2 x_3 \\ 0 & 0 & 1 & \frac 1 6 (6 x_1 + 23 x_2 - 15 x_3) \\ 0 & 0 & 0 & 18 x_1 + 56 x_2 - 36 x_3 - 6 x_4 \end{pmatrix}.$$ Then you would have a single Cartesian equation, namely $$18 x_1 + 56 x_2 - 36 x_3 - 6 x_4 = 0.$$